Question

Find the general solution of DE 2(x-1)y'=3y with a power series; check whether I can express it as an elementary function. The DE can easily be solved with another method, and I have to compare both results. I found a recursive relation of c_n+1 = c_n (n-3/2) / (n+1). I've plugged in the results for n=1, n=2, etc... and then I get c1 =-3/2 c0; c2=3 c0/(2^2.2!); c3=3 c0/2^3.3!; c4=3^2 c0/(2^4.4!); c5=5.3^2 c0/(2^5.5!); c6=7.5.3^2 c0/(2^6.6!), etc. if I plug these in y= ... The denominator can be generalized as 2^(n+1).(n+1)! but no idea how to express the numenator in general, though it's evident it gets multiplied by the next uneven number from c4 on.

Answer 1

@dumbcow

Answer 2

I know that the solution with the alternative way is y=c (x-1)^(3/2)

Answer 3

tag dumbcow long time but getting no reply, I give out my opinion instead, hihi
you have the form:
\[y =\dfrac{-3c_0}{2}+\dfrac{3c_0}{8}+\dfrac{3c_0}{48}+.....\]
and if you factor \(c_0 \) you still stuck at the first term, right?
so, let the first term out of the summation, why not??
I mean
\[y = c_0 \left(\dfrac{-3}{2}+\sum_{n=1}^\infty \dfrac{3}{2^{n+1}(n+1)}\right)\]

Answer 4

Let get @SithAndGiggles when he came online, he will fix it.

Answer 5

That seems partly right, but I don't think the numenator in the series is correct. Once you get to n=4, it becomes 5.3, and n=5 it becomes 7.5.3; n=6 becomes 9.7.5.3 and so on. But thanx so far. The last step I always kinda find a hurdle.

Answer 6

when solving it, I stop at n=2 to get c3, and then followed your solution,:) so, I am not sure about the terms you gave out.

Answer 7

\[y=c _{0}(-\frac{ 3 }{ 2} + \frac{ 3x }{2^{2}2! }+\frac{ 3.1x ^{2} }{ 2^{3} 3!}+\frac{ 3.1.3x ^{3} }{ 2^{4}4! }+\frac{ 5.3.1.3x ^{4} }{2^{5} 5!}+\frac{ 7.5.3.1.3x ^{5} }{2^{6} 6! }+...)\]
That means I can turn it into
\[y=3c _{0}(-\frac{ 1 }{ 2} + \frac{ x }{2^{2}2! }+\frac{ 1x ^{2} }{ 2^{3} 3!}+\frac{ 3.1.x ^{3} }{ 2^{4}4! }+\frac{ 5.3.1.x ^{4} }{2^{5} 5!}+\frac{ 7.5.3.1.x ^{5} }{2^{6} 6! }+...)\]

And because the uneven numbers multiplication series does not start before n=2 I can leave out the first 2 expressions?

\[y=3c _{0}(-\frac{ 1 }{ 2} + \frac{ x }{2^{2}2! }+\sum_{n=2}^{\infty}\frac{ (increasing multiplication uneven numbers)x ^{n} }{ 2^{n+1} (n+1)!}\]

Answer 8

It's actually a permutation of solely odd numbers... but I can't figure out how to write that down in a general permutation expression

Answer 9

Did some searching and googling and finally found that a k-permutation of odd numbers is called a double factorial and thus it should look like
\[y=3c _{0}(-\frac{ 1 }{ 2 }+\frac{ x }{ 2^{2}2! }+\frac{ 1!!x ^{2} }{ 2^{3}3! }+\frac{ 3!!x ^{3} }{ 2^{4} 4!}+\frac{ 5!!x ^{4} }{ 2^{5} 5!}+\frac{ 7!!x ^{3} }{ 2^{5} 5!}+...\]

An odd positive integer can be written as n=(2k-1) with k>= 1
\[(2k-1)!!=\frac{ 2k! }{ 2^{k} k!}=\frac{ (2k)^{k} }{ 2^{k} }\]

So now I'm trying to figure out how to implement that

Answer 10

If I take out another 2 and try to get the double factorial in I wonder if this is correct

\[y=\frac{ 3 }{ 2 }c _{0}(-1+\frac{ x }{ 2. 2! }+\sum_{n=2}^{\infty}\sum_{k=1}^{n}\frac{ (2k-1)!!x ^{n} }{ 2^{n} (n+1)!})\]

Answer 11

If I include the 2 and use (2k-1)!!= (2k)!/2^k. k! and n=k+1 in the above relation, I can write it down as

\[y=3 c _{0}(-\frac{ 1 }{ 2 } +\frac{ x }{ 2^{2}2! }+\sum_{n=2}^{\infty}\sum_{k=1}^{n}\frac{ (2k-1)!! x ^{k+1} }{ 2^{k+2} (k+2)!})\]
=>\[y=3 c _{0}(-\frac{ 1 }{ 2 } +\frac{ x }{ 2^{2}2! }+\sum_{n=2}^{\infty}\sum_{k=1}^{n}\frac{ (2k)! x ^{k+1} }{ 2^{k}2^{k+2} k!(k+2)!})\]
=>\[y=3 c _{0}(-\frac{ 1 }{ 2 } +\frac{ x }{ 2^{2}2! }+\sum_{n=2}^{\infty}\sum_{k=1}^{n}\frac{ (2k)! x ^{k+1} }{2^{2k+2} k!(k+2)!})\]
=>\[y=3 c _{0}(-\frac{ 1 }{ 2 } +\frac{ x }{ 2^{2}2! }+\sum_{n=2}^{\infty}\frac{ (2(n-1))! x ^{n} }{2^{2n} (n-1)!(n+1)!})\]
=> \[y=3 c _{0}(-\frac{ 1 }{ 2 } +\sum_{n=1}^{\infty}\frac{ (2(n-1))! x ^{n} }{2^{2n} (n-1)!(n+1)!})\]

I've plugged in the numbers and it seems to be correct.

Answer 12

I don't seem to be able to express that as an elementary function. And it still looks entirely different from the alternative result I got for y with the alternative (and easier) method to solve the DE. And yet I wonder whether the two expressions can be linked with a binomial?